Factor the quadratic expression completely. $12x^2+17x+6=$
Solution: Since the terms in the expression do not share a common monomial factor and the coefficient on the leading $x^2$ term is not $1$, let's factor by grouping. The expression ${12}x^2{+17}x{+6}$ is in the form ${A}x^2+{B}x+{C}$. First, we need to find two integers ${a}$ and ${b}$ such that: $\begin{cases} &{a}+{b}={B}={17} \\\\ &{ab}={A}{C}= ({12})({6})=72 \end{cases}$ We find that ${a}={8}$ and ${b}={9}$ satisfy these conditions, since ${8}+{9}={17}$ and $({8})({9})=72$. Next, we can use these values to rewrite the $x$ -term and factor by grouping. $\begin{aligned} 12x^2+17x+6&=12x^2+{8}x+{9}x+6 \\\\ &=4x(3x+2)+3(3x+2) \\\\ &=(3x+2)(4x+3) \end{aligned}$ In conclusion, $12x^2+17x+6=(3x+2)(4x+3)$